The value of `lim_(x->0)(sec4x-sec2x)/(sec3x-secx)` is

1.	The value of `lim_(x->0)(sec4x-sec2x)/(sec3x-secx)` is
Answer» `limx->0 [(sec4x-sec2x)/(sex3x-secx)]`on puttin `x=0` we get inderminate form ` 0/0 `.therefore we use l-hopital rule and differentiate numerator and denominator seprately. upon differentiating numerator and denominator seprately we get `limx->0[(4sec4xtan4x-2 sec2x tan2x)/(3sec3x tan3x-secxtanx)]` again putting value of x we get indterminate form `0/0` agan using l-hopital rule we get `limx->0[(16 sec4xtan^2 4x+16sec^3 4x-4 sec2xtan^2 2x-4sec^3 2x)/(9sex3x tan^3x+9sec^3x-secxtan^2x-sec^3 x)]` now putting value of x we get`3/2`.

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The value of `lim_(x->0)(sec4x-sec2x)/(sec3x-secx)` is

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